Question

In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS.

Show that : Area ($\Delta$ POS) = Area ($\Delta$ QOR) = $\frac{1}{2}$ Area (|| gm PQRS)

Answer 1

In given figure parallelogram PQRS, PR and QS diagonal intersect at o

We know that the diagonal intersect each other at the right angle

The line LM is parallel to PS

Then diagonals of a parallelogram divide into equal area of triangles

Then area of $\Delta$ POQ= area of $\Delta$ POS=area of $\Delta$ QOR=area of $\Delta$ SOR

OR Area of $\Delta$ POS +area $\Delta$ QOR=$\frac{1}{2}$ = area of parallelogram